Here we have a $c_o$ $=$ $3.5 \cdot 10^{-5}\;M$ solution of the strong base $Sr(OH)_2$: $Sr(OH)_2$ $\longrightarrow$ $Sr^{2+}$ $+$ $2 OH^-$ $[OH^-]$ $=$ $7.0\cdot 10^{-5}\;M$ Determine $[H_3O^+]$ !
a) $[H_3O^+]$ $ =$ $ [OH^-] $ b) $[H_3O^+]$ $ =$ $2 [OH^-] $ c) $[H_3O^+]$ $ =$ $ \frac{[OH^-]}{K_e} $ d) $[H_3O^+]$ $ =$ $\frac{[K_e]}{[OH^-]} $
Answer : d) $[H_3O^+]$ $=$ $\frac{[K_e]}{[OH^-]} $